Connecting Points in a Region $\Omega \subset \mathbb{C}$ by a Piecewise Smooth Curve

Question

A proof I read in Stein/Shakarchi's Complex Analysis involves connecting two arbitrary points in a (connected) region $\Omega \in \mathbb{C}$ by a piecewise-smooth curve $\gamma$. How do we know this is possible? All connectedness immediately implies is that there is a continuous curve connecting these two points.

I'm loosely familiar with the idea of approximating continuous functions between manifolds by smooth functions, but all that seems like overkill when the only space we care about is $\mathbb{C} \cong \mathbb{R}^2$; in any case, I highly doubt the authors expect the reader to know anything about manifolds. What would an elementary proof of this fact look like? It must be pretty simple if the authors didn't discuss it at all.

Answer 1

Given $a,b\in \Omega,$ we can actually find a polynomial curve within $\Omega$ from $a$ to $b.$ This is a consequence of the Weierstrass approximation theorem. I'll sketch the idea. First we need to tweak Weierstass a bit.

Exercise: Suppose $f:[0,1]\to \mathbb C$ is continuous. Let $\epsilon>0.$ Then there is a polynomial $p:[0,1]\to \mathbb C$ such that $|f-p|<\epsilon$ on $[0,1],$ with $p(0)=f(0),p(1)=f(1).$

I'll leave this exercise to you for the moment.

So to obtain the result, we first use the existence of a continuous $\gamma:[0,1]\to \Omega$ with $\gamma(0)=a, \gamma (1)=b.$ Then use the exercise to find a polynomial $p:[0,1]\to \mathbb C$ that is very close to $\gamma$ on $[0,1],$ and agrees with $\gamma$ at the end points. The range of $p$ will lie in $\Omega$ if "very close" is small enough.

Answer 2

Given a continuous path between two points, since the path is compact in $\mathbb{C}$, we can find a finite open covering of the path consisting of open balls, all balls contained in the region. Then choose appropriate points from the balls and connect them to form a piecewise-smooth curve (consisting of straight segments.)